`int(xsin^-1x)/sqrt(1-x^2)dx`

Explanation

To solve the integral \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx \), we can use the substitution method. Here is a step-by-step solution:

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we differentiate both sides to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{\sqrt{1 - x^2}} \implies dx = \sqrt{1 - x^2} \, dt \]

### Step 2: Express \( x \) in terms of \( t \)
From the substitution \( t = \sin^{-1} x \), we can express \( x \) as:
\[ x = \sin t \]

### Step 3: Substitute in the integral
Now substitute \( x \) and \( dx \) into the integral:
\[ \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = \int \frac{\sin t \cdot t}{\sqrt{1 - \sin^2 t}} \cdot \sqrt{1 - \sin^2 t} \, dt \]
Since \( \sqrt{1 - \sin^2 t} = \cos t \), the integral simplifies to:
\[ \int t \sin t \, dt \]

### Step 4: Integration by parts
We will use integration by parts for \( \int t \sin t \, dt \). Let:
- \( u = t \) \(\Rightarrow du = dt\)
- \( dv = \sin t \, dt \) \(\Rightarrow v = -\cos t\)

Applying integration by parts:
\[ \int t \sin t \, dt = uv - \int v \, du = -t \cos t - \int -\cos t \, dt \]
\[ = -t \cos t + \int \cos t \, dt \]
\[ = -t \cos t + \sin t + C \]

### Step 5: Substitute back for \( t \)
Now substitute back \( t = \sin^{-1} x \):
\[ = -\sin^{-1} x \cos(\sin^{-1} x) + \sin(\sin^{-1} x) + C \]
Since \( \cos(\sin^{-1} x) = \sqrt{1 - x^2} \) and \( \sin(\sin^{-1} x) = x \), we have:
\[ = -\sin^{-1} x \sqrt{1 - x^2} + x + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = -\sin^{-1} x \sqrt{1 - x^2} + x + C \]